UNIVERSITY OFDAR ES SALAAM

COLLEGE OF ENGINEERING AND TECHNOLOGY

FACULTY OF ELECTRICAL AND COMPUTER SYSTEMS ENGINEERING

DEPARTMENT OF COMPUTER SYSTEMS ENGINEERING

CS 251: INTRODUCTION TO COMPUTER ENGINEERING

ASSIGNMENT 3

GROUP MEMBERS:

SURNAME FIRST NAME DEG. PROG. REG. NUMBER

HUSSEIN TATU TE 2007-04-05286SADICK EUGENE TE 2007-04-00360

JUNE, 2009.1. Perform the following base conversions using subtraction or division-

remainder:

a) 45810 to base

b) 67710 to base5

c) 151810 to base7

d) 440110 = to base 9

2. Convert the following decimal fractions to binary with a maximum of six places to the right of the binary point:

a) 26.78125

LSB

LSB

LSB

MSB

LSB

MSB

MSB

MSB

b) 194.03125

c) 16.1240234375

3. Represent the following decimal numbers in binary using 8 – bit signed magnitude, one’s compliment and two’s compliment:

a) 77b) -42c) 119d) -107

Lsb

Msb

Lsb

Lsb

Msb

Lsb

MsbLsb

Lsb

Msb

Msb

Msb

Number 8-bit signed magnitude

One’s compliment

Two’s compliment

77

7710 = 10011012

+7710 = 010011012

7710 = 10011012

+7710=010011012

7710 = 10011012

+7710=010011012

-42

4210 = 1010102

-4210 = 101010102

4210 = 1010102

= 001010102

-4210 = 110101012

4210 = 1010102

= 001010102

Using drop down trailing zeros up to the first 1, then complimenting the rest,

-4210 = 110101102

119

11910 = 11101112

+11910=011101112

11910 = 11101112

+11910 = 011101112

11910 = 11101112

+11910 = 011101112

-107

10710 = 11010112

-10710 = 111010112

10710 = 11010112

= 011010112

-10710 = 100101002

10710 = 11010112

= 011010112

Using drop down trailing zeros up to the first 1, then complimenting the rest,

-10710 = 100101012

Lsb

Lsb

Lsb

Msb

Msb

Msb

Lsb

Msb

4. Given a (very) tiny computer that has a word size of 6 bits, what are the smallest negative numbers and the largest positive numbers that this computer can represent in each of the following representations?

a) One’s complimentGiven N = No. of bits = 6 bits

Smallest negative number = - (2N-1-1) = - (26-1-1) = - (25-1) = -31 Largest positive number = + (2N-1-1) = + (26-1-1) = + (25-1) = +31

b) Two’s compliment Smallest negative number = - (2N-1) = - (26-1) = - (25) = -32 Largest positive number = + (2N-1-1) = + (26-1-1) = + (25-1) = +31

5. Using Booths algorithm to find the multiplication of the below binary numbers

a) 1100x 101

Multiplicand = 1100;

Its 2’s compliment is 0011 +1 = 111110100

Multiplier = 0101

Record = 01010

+1 -1 +1 -1

1 1 0 0

X +1 -1 +1 -1

1 1 1 1 0 1 0 0

0 0 0 1 1 0 0

1 1 0 1 0 0

0 1 1 0 0

0 0 1 1 1 1 0 0

So

1100 X 101 = 111100

b) 10101 x 111

Multiplicand = 10101, its 2’s compliment is 01010 +1 = 01011;

Multiplier is 111; the store is 01110

+100-1

1 0 1 0 1

X +1 0 0 -1

1 1 1 0 1 0 1 1

0 0 0 0 0 0 0

0 0 0 0 0

1 0 1 0 1

1 0 0 1 0 0 1 1

So 10101 X 111 = 10010011

6. Using Booths algorithms to solve the below division.

a) 1010111 ÷101 Divisor = 1011

A register Q register

Initially 0 0 0 0 1010111

Shift left 0 0 0 1 010111

Subtract divisor 1 0 1 1

Set Qo 1 1 0 0 0101110

Shift left 1 1 0 0 0 101110

Add divisor

0 0 1 0 1

Set Qo 1 1 1 0 1 1011100

Shift left 1 1 1 0 1 1 011100

Add divisor

0 0 0 1 0 1

Set Qo 0 0 0 0 0 0 0111001

Shift left 0 0 0 0 0 0 0 111001

Subtract divisor 1 1 1 1 0 1 1

Set Qo 1 1 1 1 0 1 1 1110010

Shift left 1 1 1 1 0 1 1 1

110010

Add divisor

0 0 0 0 0 1 0 1

Set Qo 1 1 1 1 1 1 0 0

1100100

Shift left 1 1 1 1 1 1 0 0 1

100100

Add divisor

0 0 0 0 0 0 1 0 1

Set Qo 1 1 1 1 1 1 1 1 0

1001000

Shift left 1 1 1 1 1 1 1 1 0 1

001000

Add divisor

0 0 0 0 0 0 0 1 0 1

Set Qo 0 0 0 0 0 0 0 0 1 0

0010001

The Quotient is 0001 0001 and the remainder is 10.

b) 10000001÷101 -divisor= 1011

A register Q register

Initially 0 0 0 0 10000001

Shift left 0 0 0 0 1 0000001

Subtract divisor

1 1 0 1 1

Set Qo 1 1 1 0 0 00000010

Shift left 1 1 1 0 0 0 0000010

Add divisor

0 0 0 1 0 1

Set Qo 1 1 1 1 0 1 00000100

Shift left 1 1 1 1 0 1 0 0000100

Add divisor

0 0 0 0 1 0 1

Set Qo 1 1 1 1 1 1 1 00001000

Shift left 1 1 1 1 1 1 1 0 0001000

Add divisor

0 0 0 0 0 1 0 1

Set Qo 0 0 0 0 0 0 1 1 00010001

Shift left 0 0 0 0 0 1 1 0 0010001

Subtract divisor

1 1 1 1 1 0 1 1

Set Qo 0 0 0 0 0 0 0 1 0100011

Shift left 0 0 0 0 0 0 0 1 0

100011

Subtract divisor

1 1 1 1 1 1 0 1 1

Set Qo 1 1 1 1 1 1 1 0 1

1000110

Shift left 1 1 1 1 1 1 1 0 1 1

000110

Add divisor

0 0 0 0 0 0 0 1 0 1

Set Qo 0 0 0 0 0 0 0 0 0 0

0001100

The Quotient is 1100, with a remainder of 0

7.

Sign=1digit Exponent = 5 digits Mantissa =8 digits

Exponent bias = 16.

The number 100.010 =01100100.0 = 1.1001 X 106

Mantissa = 10010000

Exponent = 6+16 = 22 = 10110

Sign = 0

The number 0.25 = 0.01=1.0X10-2

Mantissa = 01000000

Exponent = 16+-2=14 = 1110

Sign =0

8. Why do we usually store floating point numbers in the normalized form? What are the advantages of using a bias as opposed to adding a sign bit to the exponent? Answer:

The reason of using the normalized form floating point numbers is that they tend to allow very large and very small numbers to be presented using only few digits at the expense of precision

Used in many application to represent numbers since most of them they are not capable of using integers

Advantages of using bias as opposed to sign bit is Easy to present the floating point numbers

0 10110 10010000

0 01110 01000000

Accuracy in the floating numbers representation

9. Given that the ASCII code for A is 1000001, what is the ASCII code for J?

Given that the EBCDIC code for A is 11000001, what is the EBCDIC code for J?

(a) A=1000001

To change A into hexadecimal number:

A=0100|0001=4116

From A to J we have 9 characters, so add 9 in A to get J

9 in binary is 1001

J=A+9=1000001+1001=1001010

J=1001010

Hence the value of J is 1001010

(b) A=11000001

To change A into hexadecimal number:

A=1100/0001=121

From A to J we have 9 characters, so add 9 in A to get J

9 in binary is 1001

J=A+9=11000001+1001=11001010

J=11001010

Hence the value of J is 11001010

10.

Assume a 24 –bit word on a computer .In these 24 bits we wish to represent the value 295.

(a) If our computer uses even parity ,how would the computer represent the decimal value 295?

(b) If our computer uses 8-bit ASCII and even parity ,how would be computer represent the string 295

(c) If our computer uses packed BCD ,how would the computer represent the number +295

(a) To represent the decimal value 295

Change 295 into binary

29510=100100111

If we use even parity 295 will be 000000000000001100100111

(b) If we use 8-bit ASCII and even parity

To change 295 into binary

295=100100111

If we use even parity 295 will be 000000000001000100100111

(c) If a computer uses packed BCD

295 in BCD=001010010101

BCD using 24 bits 295 will be equal to 000000000000001010010101

11.

Decode the following ASCII message, assuming 7-bit ASCII character and no parity.

1001010 1001111 1001000 1001110 0100000 1000100 1000101.

To change each binary into hexadecimal

1001010=4A, 1001111=4F, 1001000=48, 1001110=4E, 0100000=20, 1000100=44,

1000101=45

To change each hexadecimal into its equivalent character

4A=J ,4F=0 ,48=H ,4E=N ,20=BLANK ,44=D ,45=E

When 1001010 1001111 1001000 1001110 0100000 1000100 1000101 is decoded it will be equal to JOHN DE

12.

Find the quotients and remainders for the following division problems modules.

(a) 10101112÷11012

(b) 10111112÷111012

Solution.

0000110a/ 1101 -1101 10001 - 1101 1001 0000 1001

Quotient is 110 and remainder is 1001

b/ 0000011 11101 -11101 100101 -11101 01000

Quotient is 11 and remainder is 10000

13.Convert the IEEE single precision Floating Point numbers to the decimal values

Answer:

The number i

The number is

14.

In order to print out number 3.142, which five ASCII characters must be sent to the printer?

Answer:

15.

Describe IEEE 754 Floating Point Standard of representing fraction numbers.

Answer:

This is the standard format for representing binary floating point numbers introduced by IEEE.The number is represented by using three fields,sign field,exponent field and mantissa field.The three fields are represented in the form of where “ f” is the value of the fraction field and “e” is the

biased exponent.

Example:

In single precision

In tabular form:

S e m

0 0000...........................0000

1 bit 8 bits 32 bits

In double precision

e

m

S e m

0 0000........................................0000

1 bit 11 bits 64 bits

16.

Calculate the decimal value represented by the binary number 10101 using

a) Signed magnitude c) Two’s complement

b) One’s complement d) Excess representation

Answer:

a) Using Signed magnitude

b) Using One’s complement

20

c) Using Two’s complement

hence the number is 9

d) Using Excess representation

16. Calculate the decimal value represented by the binary number 10101 using

(a) signed magnitude(b)one’s compliment(c) two’s compliment(d)excess representation

(a) signed magnitudeIn signed magnitude the msb indicates the sign of the decimal number

then 1 in 101012 indicates that the number is negative of magnitude

= 20 +22 = 1+4 = -510

b) one’s compliment

In one’s complement 101012 become 010102

= 21 +23

=1010

(c) two’s complimentIn 2’s complement 101012 become 010112

=20 +21 +23 =1110

(d)excess representation 101012 its bias = 25-1 =16=100002

2’s complement = 010112 Add bias = 16 = 1000Answer = 110112

The number is positive since msb is 1,=20 +21+23

=1110

17.What is the Booth’s Algorithm? State the steps required for multiplying two signed binary numbers. When multiplying -6 times -5, what is the sequence of values in the (P, A) register pair?

Booth algorithm- Is the technique used to speed up the multiplication operation based on the observation that the larger the number of consecutive zeros and ones, the fewer partial products that have to be generated.

Steps required performing multiplication1) If x is the count of bits of the multiplicand, and y is the count of bits of the

multiplier : 2) Draw a grid of three lines, each with squares for x + y + 1 bits. Label the

lines respectively A (add), S (subtract), and P (product). 3) In two's complement notation, fill the first x bits of each line with : 4) A: the multiplicand 5) S: the negative of the multiplicand 6) P: zeroes 7) Fill the next y bits of each line with : 8) A: zeroes 9) S: zeroes

i. P: the multiplier

10) Fill the last bit of each line with a zero. 11) Do both of these steps |y| (the Absolute Value of y) times : 12) If the last two bits in the product are... 13) 00 or 11: do nothing. 14) 01: P = P + A. Ignore any overflow. 15) 10: P = P + S. Ignore any overflow. 16) Arithmetically shift the product right one position. 17) Drop the last bit from the product for the final result.

18) Sequence value is of (P, A) in register pair for -6*-519) A = 0011 0000 0 20) S = 1101 0000 0 21) P = 0000 1100 0 22) Perform the loop four times :

23) P = 0000 1100 0. The last two bits are 00. 24) P = 0000 0110 0. A right shift. 25) P = 0000 0110 0. The last two bits are 00. 26) P = 0000 0011 0. A right shift. 27) P = 0000 0011 0. The last two bits are 10. 28) P = 1101 0011 0. P = P + S. 29) P = 1110 1001 1. A right shift. 30) P = 1110 1001 1. The last two bits are 11. 31) P = 1111 0100 1. A right shift.

a. The product is 1111 0100, which is 30

18.

Design a full subtractor using only the NOR gates

Truth table

X Y Bin D Bout

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

0

1

1

0

1

0

0

1

0

1

1

1

0

0

0

1

Logic expressions:

Bout

D

BinYX

19. Design a full adder circuit using only the NAND gates.

Cin

Sum

Carry

BA